Solve {l}{64a-8b+c=8}{64a+8b+c=8}{c=11}

$$\left\{ \begin{array} { l } { 64 a - 8 b + c = 8 } \\ { 64 a + 8 b + c = 8 } \\ { c = 11 } \end{array} \right.$$

Solve for a, b, c

$$a=-\frac{3}{64}=-0.046875$$

$$b=0$$

$$c=11$$

Short Steps Using Substitution

Reorder the equations.

$$c=11 64a+8b+c=8 64a-8b+c=8$$

Substitute 11 for c in the second and third equation.

$$64a+8b+11=8 64a-8b+11=8$$

Solve these equations for b and a respectively.

$$b=-8a-\frac{3}{8} a=\frac{1}{8}b-\frac{3}{64}$$

Substitute -8a-\frac{3}{8} for b in the equation a=\frac{1}{8}b-\frac{3}{64}.

$$a=\frac{1}{8}\left(-8a-\frac{3}{8}\right)-\frac{3}{64}$$

Solve a=\frac{1}{8}\left(-8a-\frac{3}{8}\right)-\frac{3}{64} for a.

$$a=-\frac{3}{64}$$

Substitute -\frac{3}{64} for a in the equation b=-8a-\frac{3}{8}.

$$b=-8\left(-\frac{3}{64}\right)-\frac{3}{8}$$

Calculate b from b=-8\left(-\frac{3}{64}\right)-\frac{3}{8}.

$$b=0$$

The system is now solved.

$$a=-\frac{3}{64} b=0 c=11$$